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3 years ago

An emerald gemstone is an impure form of the mineral beryl, Be3Al2Si16O18. What is the percentage of silicon in the mineral?

Chemistry

1 answer:

solong [7]3 years ago

4 0

Answer:

b. 54.9%

Explanation:

An emerald gemstone has the formula Be₃Al₂Si₁₆O₁₈. We can find the mass of each element in 1 mole of Be₃Al₂Si₁₆O₁₈ by multiplying the molar mass of the element by its atomicity.

Be: 3 × 9.01 g = 27.03 g

Al: 2 × 26.98 g = 53.96 g

Si: 16 × 28.09 g = 449.4 g

O: 18 × 16.00 g = 288.0 g

Total mass = 818.4 g

The mass percentage of silicon is:

(449.4 g / 818.4 g) × 100% = 54.91%

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4 years ago

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The statement that best describes the effect of low ionization energies and low electronegativities on metallic bonding is the first one - the valence electrons are easily delocalized.
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3 years ago

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If the theoretical yield of RX is 56.0 g , what is the percent yield ?

maks197457 [2]

the complete question in attached figure

Let
x------------------- > actual yield
y------------------- > theoretical yield
z------------------- > percent yield
we have that
z=x/y

we know
x=47 g
y=56 g

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the answer is the option C 83.9%

4 0

3 years ago

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of

zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

CH3COOH + KOH ↔ CH3COONa + H2O

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

(C*V)acid = (C*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH = 0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0

3 years ago

Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupies a volume

pishuonlain [190]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

mass = 0.508 g, Volume = 0.175 L

Temperature = (25 + 273) K = 298 K, P = 1 atm

As per the ideal gas law, PV = nRT.

where, n = no. of moles = $\frac{mass}{\text{molar mass}}$

Hence, putting all the given values into the ideal gas equation as follows.

PV = $\frac{mass}{\text{molar mass}} \times RT$

1 atm \times 0.175 L = $\frac{0.508 g}{\text{molar mass}} \times 0.0821 L atm/ K mol \times 298 K$

= 71.02 g

As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of $Cl_{2}$ is 70.8 g/mol or 71 g/mol (approx).

Thus, we can conclude that the gas is most likely chlorine.

4 0

3 years ago